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  • Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Q: 4         Find a point on the x-axis, which is equidistant from the points  (7,6) and (3,4).

Answers (1)

best_answer

Point is on the x-axis, therefore, y coordinate is 0
Let's assume the point is (x, 0)
Now, it is given that the given point (x, 0) is equidistance from point  (7, 6) and (3, 4)
We know that
Distance between two points is given by
D = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|
Now,
D_1 = |\sqrt{(x-7)^2+(0-6)^2}|= |\sqrt{x^2+49-14x+36}|= |\sqrt{x^2-14x+85}|
and
D_2 = |\sqrt{(x-3)^2+(0-4)^2}|= |\sqrt{x^2+9-6x+16}|= |\sqrt{x^2-6x+25}|
Now, according to the given condition
D_1=D_2
|\sqrt{x^2-14x+85}|= |\sqrt{x^2-6x+25}|
Squaring both the sides
x^2-14x+85= x^2-6x+25\\ 8x = 60\\ x=\frac{60}{8}= \frac{15}{2}
Therefore, the point is ( \frac{15}{2},0)
 

Posted by

Gautam harsolia

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